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Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|2x+3\right|-2\left|4-x\right|=5\)
\(\Rightarrow\left|2x+3\right|-\left|8-2x\right|=5\)
\(\Rightarrow\left|2x+3\right|=5+\left|8-2x\right|\)
+) \(TH_1:2x+3\ge0\Rightarrow2x\ge3\Rightarrow x\ge\frac{3}{2}\)
\(2x+3=5+8-2x\)
\(\Rightarrow2x+2x=-3+13\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}.\)
+) \(TH_2:2x+3< 0\Rightarrow2x< -3\Rightarrow x< \frac{-3}{2}\)
\(-2x-3=5+8-2x\)
\(\Rightarrow-2x+2x=3+13\)
\(\Rightarrow0=16\) (vô lí)
Vậy \(x=\frac{5}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: A(x)=0
=>5x-10-2x-6=0
=>3x-16=0
=>x=16/3
b: B(x)=0
=>5x^2-125=0
=>x^2-25=0
=>x=5 hoặc x=-5
c: C(x)=0
=>2x^2-x-3=0
=>2x^2-3x+2x-3=0
=>(2x-3)(x+1)=0
=>x=3/2 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Ta có: 5x=4y+2x
\(\Leftrightarrow3x=4y\)
\(\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{-56}{7}=-8\)
Do đó: x=-32; y=-24
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Ta có: 5x=4y+2x
\(\Leftrightarrow3x=4y\)
hay \(\dfrac{x}{4}=\dfrac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{-56}{7}=-8\)
Do đó: x=-32; y=-24
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m\left(x\right)+h\left(x\right)=g\left(x\right)-5\)
\(\Leftrightarrow m\left(x\right)=g\left(x\right)-h\left(x\right)-5\)
\(\Leftrightarrow m\left(x\right)=4x^2+3x+1-3x^2+2x+3-5\)
\(\Leftrightarrow m\left(x\right)=x^2+5x-1\)
\(B=\frac{2x-1}{x+5}\)
\(\Rightarrow B=\frac{2x+10-11}{x+5}\)
\(\Rightarrow B=2-\frac{11}{x+5}\)
Để \(B\in Z\)
\(\Rightarrow x+5\in\left(1;11;-1;-11\right)\)
\(\Rightarrow x\in\left(-4;6;-6;-16\right)\)
\(B=\frac{2x-1}{x+5}=\frac{2\left(x+5\right)-9}{x+5}=\frac{2\left(x+5\right)}{x+5}-\frac{9}{x+5}=2-\frac{9}{x+5}\)
Rút gọn biểu thức B ta được \(B=2-\frac{9}{x+5}\)hoặc \(B=\frac{2x}{x+5}-\frac{1}{x+5}\)
ĐKXĐ của B : \(x+5\ne0\Rightarrow x\ne-5\)
P/S : Đề ko có yêu cầu gì hết nên mình chỉ làm vậy thôi